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Electron-transfer reduction of selected alcohols with alkalide K, K(15-crown-5)2 via organometallic intermediates
Authors:Zbigniew Grobelny  Adalbert Maercker  Janusz Kasperczyk  Holger Frey
Institution:a Institute of Materials Science, University of Silesia, Bankowa 12, 40-007 Katowice, Poland
b Institut für Organische Chemie, Universität Siegen, 57068 Siegen, Germany
c Department of Analytical and General Chemistry, Silesian University of Technology, 44-101 Gliwice, Poland
d Chair of Molecular Biology, Biochemistry and Biopharmacy, Medical University of Silesia, 41-200 Sosnowiec, Poland
e Institute of Chemistry, University of Silesia, 40-007 Katowice, Poland
f Institut für Organische Chemie, Johannes-Gutenberg-Universität, 55099 Mainz, Germany
Abstract:The course of the reaction of alkalide K, K+(15-crown-5)21 with selected alcohols depends on the kind of alcohol and the mode of substrate delivery. In the case of methanol, potassium methoxide formed initially undergoes destruction at the excess of 1. It results in potassium oxide and methylpotassium. The latter opens the crown ether ring giving potassium tetraethylene glycoxide vinyl ether and methane. A similar course of the process is observed for propanol. Potassium glycidoxide is the main product formed in the reaction of 1 with glycidol. Its oxirane ring is opened at the excess of 1. Organopotassium alkoxides, i.e., potassium potassiomethoxide and dipotassium potassiopropane-1,2-dioxide are intermediate products of this reaction. They react then with the crown ether. Potassium methoxide, potassium enolate of acetaldehyde, dipotassium propane-1,2-dioxide and potassium tetraethylene glycoxide vinyl ether are the final products of this process.
Keywords:Methanol  Glycidol  Propanol  Alkalide  Potassium anions  Electron-transfer reduction  Organopotassium intermediates
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